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STRUCTURE OF Tb-159
By Prof Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of electromagnetic laws in favor of wrong theories which could not lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for discovering the nuclear force and structure by applying the well-established laws of electromagnetism. (See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). ' ' STRUCTURES OF TERBIUM Naturally occurring terbium (Tb) is composed of 1 stable isotope, Tb-159. 36 radioisotopes have been characterized, with the most stable being Tb-158 with a half-life of 180 years, Tb-157 with a half-life of 71 years, and Tb-160 with a half-life of 72.3 days Comparing the structures of Terbiumb of 65 protons (odd number ) with those of Gd of 64 protons (even number) we see that the Terbiumb has only one stable isotope because it breaks the high symmetry of structures of Gd . (See my STRUCTURE OF Gd-154...Gd-160 ). After a careful analysis of this comparison I discovered that in the structure of Gd the blank positions of the extra alpha particles on the right and on the left side of the central parallelepiped are complete. So the additional vertical p65n65 along with the p39n39 of the down square come to fill the blank positions in front of the central parallelepiped and behind it with 2(n). However the p39n39 changes the spin from S=-1 to S = 0 giving S =+1. In the diagram of Terbium the p65n65 is not shown but you can see the n65 near the p49 by using the top view of the third plane , because the n65p65 as a vertical pn system makes a rectangle with the p49n49. Also at the same top view you can see the p39 because the p39n39 makes with n50p50 the symmetrical vertical rectangle. Here you can also see the blank positions of 2(n) formed by p39. So under these symmetrical arrangements the number N of blank positions is given by The p37n37 gives 3n of negative spins The up horizontal squares gives 4n of positive spins The first and the sixth plane give 4(n) of weak bonds with opposite spins. The second and the fifth plane give 4{n} with three bonds per neutron and 8n. The third and the fourth plane give 4(n) of weak bonds with opposite spins. The two symmetrical alpha particles give at the same planes 4(n) of opposite spins. The two symmetrical rectangles at the same planes give 4(n) of opposite spins That is N = 4{n} +15n + 16(n) = 35 blank positions able to receive 18 extra neutrons of positive spins and 17 extra neutrons of negative spins . ' ' STRUCTURE OF Tb-159 WITH S = +3/2 Since the Tb-159 of 29 extra neutrons has S =+3/2 we conclude that it has 15 extra neutrons of positive spins and 14 extra neutrons of negative spins. That is S = +1 + 15(+1/2) + 14(-1/2) = +3/2 Especially it has 2{n} +8n + 5(n) of positive spins and 2{n} + 7n + 3(n). DIAGRAM OF TERBIUM-130 FORMING 35 BLANK POSITIONS Here the additional pn systems as vertical p63n63 and n634p64 with S = 0 are shown near the n62p62 and p61n61 respectively. However the additional n65p65 along with the symmetrical p39n39 are not shown. Note that using the top view of the third horizontal plane one can see the n65 near thep49 and the p39 near then50 along with 2(n). In the diagram you can see the p47n47 along with the p48n48 make two inner symmetrical alpha particles of opposite spins . But you cannot see the p49n49, the n52p52 of the third alpha particle and the n50p50 and the p51n51 of the fourth alpha particle. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. In the same way the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. n40......p40........n ' n......... p38.......n38 H. Square with n' ' n31………p12.........n12.......p32' ' p31........n11.........p11…… n32 Sixth h. plane' ' n........p29.........n10.........p10…… n30 ' ' n29…… p9..........n9 …….p30.........n Fifth h. plane' ' n61....p47.......n27.........p8..........n8.........p28........... n48......p62' ' p64....n45...........p27........n7.........p7........n28..........p46...........n63 Fourth h. plane ' ' p61......n47..........p25.........n6.........p6..........n26...........p48.....n62' ' n64....p45..........n25…….p5..........n5……….p26.......n46 .........p63 Third h. plane' ' n23………p4........n4………….p24..............n' ' n......p23…….....n3………....p3…….n24 Second h. plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22] First h. plane' ' n.........p37......n37 p37n37 with n' ' ' TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' Here you see the 2(n) of weak horizontal bonds ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22' ' n33.......p33..... (n)' ' TOP VIEW OF THE SECOND HORIZONTAL PLANE' Here we have 2{n} +4n and the same situation occurs at the fifth horizontal plane. ' n' ' n14.......p14........{n}' ' n23.......p4.........n4.........p24..........n ' ' n.......p23........n3........p3.........n24' ' {n}...... p13......n13 ' ' n' TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS ' Here the first 2(n) of horizontal bonds fill the blank positions of the central parallelepiped, while the second 2(n) of horizontal bonds are formed by the additional alpha particles. Also the p39 near n50 gives 1(n) near the p58. Using this top view of the third plane you can see the following characteristics of the fundamental shapes formed by the nucleons of the central parallelepiped as The p5n5 and n6p6 create the small horizontal square of Mg-24 for creating the central parallelepiped of the alpha particle nuclei. The n15p15 and p16n16 create the first small horizontal rectangle. The p25n25 and p26n26 create the second small horizontal rectangle. The p41, n42, n43 and p44 make the great horizontal square of the great central parallelepiped. The p45, n46, n47 and p48 form the first great horizontal rectangle. The p49, n50, p51 and n52 form the second great horizontal rectangle. ' ' ' (n) p39 (n) ' (n)........p58....... n50.....p51......n60 ' ' (n) p53........n42........p16......n16......p44.......n54' ' p61 n47........p25........n6........p6........n26.......p48 n62' ' n64 p45........n25........p5........n5........p26...... n46 p63 ' ' n55........p41.......n15.......p15.......n43......p56 (n)' ' n57.......p49.......n52......p59........(n)' ' n65 ' TOP WIEW OF THE UP HORIZONTAL SQUARE Here the 2n near p38 have strong bonds with p31 and p35 respectively. Whereas the 2n near p40 have the strong bonds with p32 and p36 respectively. ' n' ' n40......p40.......n ' n.....p38.......n38 ' n' Category:Fundamental physics concepts